Question: $\sum\limits_{n=1}^{\infty } \dfrac{(n+5)(x)^n}{n^2\cdot5^n}$ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-5 \le x<5$ (Choice B) B $-10 < x \le 0$ (Choice C) C $-10 \le x<0$ (Choice D) D $-5 < x \le5$
Explanation: We use the ratio test. For $x\neq0$, let $a_n=\dfrac{(n+5)(x)^n}{n^2\cdot5^n}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x}{5}\right| $ The series converges when $\left| \dfrac{x}{5}\right|<1$, which is when $-5<x<5$. Now let's check the endpoints, $x=-5$ and $x=5$. Letting $x=-5$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty }\dfrac{(n+5)(-5)^n}{n^2\cdot5^n} &=\sum\limits_{n=1}^{\infty }\dfrac{(n+5)(-1)^n5^n}{n^2\cdot5^n}\\\\ &=\sum\limits_{n=1}^{\infty }(-1)^n\cdot \dfrac{(n+5)}{n^2} \end{aligned}$ By the alternating series test, we know this series converges. Letting $x=5$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{(n+5)(5)^n}{n^2\cdot5^n} &=\sum\limits_{n=1}^{\infty } \dfrac{(n+5)}{n^2}\\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{1}{n} + \dfrac{5}{n^2} \end{aligned}$ By direct comparison with the harmonic series, we know this series diverges. In conclusion, the interval of convergence is $-5 \le x<5$.